如何使用 R 根据 group_by cols 和时间戳计算平均响应时间和总响应时间？

Question

我有一个看起来像这样的表（1 个示例 - 总共 200 万行）：

tweet_id |  id         |  group   |                created_at         | tweet                                                 |  response_tweet_id
1           sprintcare    Support      Tue Oct 31 22:10:47 +0000 2017   @115712 I understand. I would like to assist you.        2
2           115712        Customer     Tue Oct 31 22:11:45 +0000 2017   @sprintcare and how do you propose we do that?           NA
3           115712        Customer     Tue Oct 31 22:08:27 +0000 2017   @sprintcare I have sent several private messages.        1
4           sprintcare    Support      Tue Oct 31 21:54:49 +0000 2017   @115712 Please send us a Private Message.                3
5           115712        Customer     Tue Oct 31 21:49:35 +0000 2017   @sprintcare I did.                                       4
6           sprintcare    Support      Tue Oct 31 21:46:24 +0000 2017   @115712 Can you please send us a private message.        5,7
8           115712        Customer     Tue Oct 31 21:45:10 +0000 2017   @sprintcare is the worst customer service                9,6,10
11          apple         Support      Tue Oct 31 22:10:35 +0000 2017  @115713 This is saddening to hear.                        NA
12          115713        Customer     Tue Oct 31 22:04:47 +0000 2017   @apple My phone is not working. Help!               11

引用第一行时，我们可以看到响应 tweet_id 1（标记为 2）时第一个支持 tweet 响应的时间。

理想情况下 - 我想计算每条响应推文需要多长时间 - 从支持到客户订购。

我想计算两个值：

1. 向支持人员发出第一条推文之间的响应时间（tweet_id、response_tweet_id 和 created_at）。对于第一行 - 时差是： 2017 年 10 月 31 日星期二 22:10:47 +0000 - 2017 年 10 月 31 日星期二 22:08:27 +0000 = 00:02:20。

2. 从第一条推文从支持到给每个客户的最后一条推文之间的总响应时间。在下面的示例中 - 它本质上是相对于支持时间发出的第一条推文，以及在下一个唯一 ID 发挥作用之前发出的最后一条推文。

根据每个组（100 多家独特的公司），所需的输出如下所示：

id         | group  | Avg_response_time_per_tweet (in minutes) | Total_avg_response_time (in minutes)
sprintcare   Support  ####                                       ####
apple        Support  ####                                       #### 

Answer 1

library(lubridate)
library(dplyr)
library(tidyr)
df %>% 
    separate(created_at, c("Day_name", "Month", "Day", "Hour", "Minute", "Second", "X", "Year")) %>% 
    type.convert(as.is = TRUE) %>% 
    mutate(Month = match(Month, month.abb)) %>% 
    mutate(created_at = make_datetime(Year, Month, Day, Hour, Minute, Second), .keep="unused") %>% 
    group_by(id, group) %>% 
    summarise(Avg_response_time = mean(difftime(max(created_at), min(created_at))))

  id         group    Avg_response_time
  <chr>      <chr>    <drtn>           
1 115712     Customer 26.58333 mins    
2 sprintcare Support  24.38333 mins  

Answer 2

这不是一个真正高效的方法，但它应该会产生预期的结果。

library(dplyr)
library(stringr)

dat %>% 
  mutate(id_at = str_extract(tweet, "\\d+"),
         global_id = ifelse(is.na(id_at), id, id_at)) %>% 
  group_by(global_id) %>% 
  group_modify(~ .x %>%
                 rowwise %>% 
                 mutate(response_time = if(group == "Support") {
                   filter(.x, created_at < .env$created_at, group == "Customer") %>% 
                     slice_max(created_at, n = 1) %>% pull(created_at)
                 } else NA)
  ) %>% 
  summarise(id = first(id), Total_avg_response_time = difftime(max(created_at), min(created_at)),
            Avg_response_time_per_tweet = mean(difftime(created_at, response_time), na.rm = TRUE))

#> # A tibble: 1 x 4
#>   global_id id         Total_avg_response_time Avg_response_time_per_tweet
#>   <chr>     <chr>      <drtn>                  <drtn>                     
#> 1 115712    sprintcare 26.58333 mins           2.933333 mins

# data used
dat <- read.table(text = "tweet_id   id             group                  created_at            tweet
                  1           sprintcare    Support      'Tue Oct 31 22:10:47 +0000 2017'   '@115712 I understand. I would like to assist you.'
                  2           115712        Customer     'Tue Oct 31 22:11:45 +0000 2017'   '@sprintcare and how do you propose we do that'
                  3           115712        Customer     'Tue Oct 31 22:08:27 +0000 2017'   '@sprintcare I have sent several private messages and no one is responding as usual'
                  4           sprintcare    Support      'Tue Oct 31 21:54:49 +0000 2017'   '@115712 Please send us a Private Message so that we can further assist you.'
                  5           115712        Customer     'Tue Oct 31 21:49:35 +0000 2017'   '@sprintcare I did.'
                  6           sprintcare    Support      'Tue Oct 31 21:46:24 +0000 2017'   '@115712 Can you please send us a private message, so that I can gain further details about your account?'
                  7           115712        Customer     'Tue Oct 31 21:45:10 +0000 2017'   '@sprintcare is the worst customer service'",
                  header = TRUE)

dat <- dat %>% 
  mutate(created_at = strptime(created_at, "%a %b %d %H:%M:%S %z %Y"))

^{由reprex package (v0.3.0) 于 2021 年 9 月 15 日创建}